update: auto commit
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\begin{frame}{Appendix}{Why can't we fully automate the workflow using Mathematica?}
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The correct way is to check the geometric multiplicity for each eigenvalues. Say we get eigenvalues $\lambda_1, ..., \lambda_k$ with the algebraic multiplicity $m_1, ..., m_k$. Under the assumption that all these eigenvalues are different from each other, we check $\dim\ker(A - \lambda_i I)$ for each $i$. $\dim\ker(A - \lambda_i I) \leq m_i$, so we just have to find the condition for $\dim\ker(A - \lambda_i I) \geq m_i$. $\dim\ker(A - \lambda_i I) = n - \rank(A - \lambda_i I)$, so we check $\rank(A - \lambda_i I) \leq n - m_i$. The criterion is that any $(n-m_i+1)\times(n-m_i+1)$ minor of $A - \lambda_i I$ is zero.
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Nice, this is a computable condition! But physically we expect most modes to have the same characteristic speed $\shift{^\xi}$, resulting a algebraic multiplicity $\sim 22$, while the size of matrix is 55, then the number of minors to check is
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Nice, this is a computable condition! But physically we expect most modes to have the same characteristic speed $\shift{^\xi}$, resulting a algebraic multiplicity $\sim 22$, while the size of matrix is 55, then the number of minors to check for $\lambda = \shift{^\xi}$ is
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\begin{equation*}
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\binom{55}{34}^2 = 708507400701023142362023505625 = 7.085\times 10^{29}.
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\end{equation*}
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